gle contained under A and C, is equal to the Rectangle under B and D; but the Rectangle under B and D is equal to the Square of D, for B is equal to D. Wherefore the Rectangle contained under A, C, is equal to the Square of B.

And if the Rectangle contained under A, C, be equal to the Square of B; I say, as A is to B, fo is B to C.

For the fame Conftruction remaining, the Rectangle contained under A and C is equal to the Square of B; but the Square of B is the Rectangle contained under B, D, for B is equal to D, and the Rectangle contained under A, C, fhall be equal to the Rectangle contained under B, D. But if the Rectangle contained under the Extremes, be equal to the Rectangle contained under the Means, the four Right Lines +16 of this. fhall be + Proportionals. Therefore A is to B as D is to C; but B is equal to D. Wherefore A is to B, as B is to C. Therefore, if three Right Lines be proportional, the Rectangle contained under the Extremes, is equal to the Square of the Mean; and if the Rectangle under the Extremes, be equal to the Square of the Mean, then the three Right Lines are proportional; which was to be demonftrated.

23. I.

PROPOSITION XVIII.

PROBLEM.

Upon a given Right Line, to defcribe a Right-lined Figure fimilar, and fimilarly fituate to a Rightlined Figure given.

ET AB be the Right Line given, and CE the Right-lined Figure. It is required to defcribe upon the Right Line AB a Figure fimilar, and fimilarly fituate to the Right-lined Figure CE.

Join DF, and make* at the Points A and B, with the Line AB, the Angles GAB, ABG, each equal to the Angles C and CDF. Whence the other An+Cor. 32. 1. gle CFD is + equal to the other Angle AGB; and fo the Triangle FCD is equiangular to the Triangle GAB; and confequently, as FD is to GB, fo is of this. FC to GA; and fo is CD to AB. Again, make the Angles BGH, GBH, at the Points B and G,

* II. 5.

with the Right Line BG, each equal to the Angles EFD, EDF; then the remaining Angle at E, is †† Cor. 32. 1. equal to the remaining Angle at H. Therefore the Triangle FDE, is equiangular to the Triangle GBH; and confequently, as FD is to GB, fo is 1 FE tot 4 of this. GH; and fo ED to HB. But it has been proved that FD is to GB, as F C is to GA, and as CD to AB. And therefore as FC is to AG, fo is* CD to AB; and fo FE to GH; and fo ED to HB. And because the Angle CFD is equal to the Angle AGB; and the Angle DFE equal to the Angle BGH; the whole Angle CFE fhall be equal to the whole Angle AGH. By the fame Reason, the Angle CDE is equal to the Angle ABH; and the Angle at C equal to the Angle A; and the Angle E equal to the Angle H. Therefore the Figure A H is equiangular to the Figure CE; and they have the ́ Sides about the equal Angles proportional. Confequently, the Right-lined Figure AH will be + fimilart Def. 1. to the Right-lined Figure CE. Therefore there is of this. defcribed upon the given Right Line AB, the Rightlined Figure AH fimilar, and fimilarly fituate to the given Right-lined Figure CE; which was to be done.

PROPOSITION XIX.

THEOREM.

Similar Triangles are in the duplicate Proportion of their homologous Sides.

ET ABC, DEF, be fimilar Triangles, having the Angle B equal to the Angle E; and let AB be to BC as DE is to EF, fo that BC be the Side homologous to EF. I fay, the Triangle ABC, to the Triangle DEF, has a duplicate Proportion to that of the Side BC to the Side EF.

For take BG a third Proportion to BC and EF;* 11 of this. that is, let BC be to EF, as EF is to BG, and join GA. Then because AB is to BC, as DE is to EF; it fhall be (by Alternation) as A B is to DE, fo is BC to EF; but as BC is to EF, fo is EF to BG. Therefore as AB is to DE, fo is +EF to BG; confe- + 11. 5. quently, the Sides that are about the equal Angles of

M 3

the

the Triangles ABG, DEF, are reciprocal: But those Triangles that have one Angle of the one, equal to on e Angle of the other; and the Sides about the equal An15 of this gles reciprocal, are equal. Therefore the Triangle ABG, is equal to the Triangle DEF; and because BC is to EF, as EF is to BG, and if three Right Lines Def. 1o. 5. be proportional, the firft has a duplicate Proportion to the third, of what it has to the fecond. BC to BG fhall have a duplicate Proportion of that which BC has to EF; and as BC is to BG, so is the Triangle ABC to the Triangle ABG, whence the Triangle ABC bears to the Triangle ABG a duplicate Proportion to what BC doth to EF; but the Triangle ABG is equal to the Triangle DEF. Therefore the Triangle ABC, to the Triangle DEF, shall be in the duplicate Proportion of that which the Side BC has to the Side EF. Wherefore fimilar Triangles are in the duplicate Proportion of their homologous Sides; which was to be demonstrated.

Coroll. From hence it is manifeft, if three Right Lines be proportional, then as the first is to the third, fo is a Triangle made upon the firft to a fimilar, and fimilarly defcribed Triangle upon the fecond: Because it has been proved, as CB is to BG, fo is the Triangle ABC to the Triangle ABG, that is, to the Triangle DEF; which was to be demonftrated.

PROPOSITION XX.

THEOREM.

Similar Polygons are divided into fimilar Triangles, equal in Number, and homologous to the Wholes; and Polygon to Polygon, is in the duplicate Proportion of that which one homologous Side has to the other.

ET ABCDE, F GHKL, be fimilar Polygons; and let the Side AB be homologous to the Side FG. I fay, the Polygons ABCDE, FGHKL, are divided into equal Numbers of fimilar Triangles, and homologous to the Wholes; and the Polygon ABCDE, to the Polygon FGHKL,, is in the duplicate Proportion of that which the Side AB has to the Side FG.

For let BE, EC, GL, LH, be joined.

Then

*

Then because the Polygon ABCDE is fimilar to the Polygon FGHKL, the Angle BAE is equal to the Angle GFL; and BA is to AE as GF is to FL. Now fince ABE, FGL, are two Triangles, having one Angle of the one equal to one Angle of the other, and the Sides about the equal Angles pro-` portional; the Triangle ABE will be equiangular * 6 of this. to the Triangle FGL; and alfo fimilar to it. Therefore the Angle ABE, is equal to the Angle FGL; but the whole Angle ABC is + equal to the whole † De 1. of this. Angle FGH, because of the Similarity of the Polygons. Therefore the remaining Angle EBC is equal to the remaining Angle LGH: And fince (by the Similarity of the Triangles ABE, FGL) as EB is to BA, fo is LG to GF And fince alfo by the Similarity of the Polygons) AB is to BC, as FG is to GH; it fhall be by Equality of Proportion, as 22. EB is to BC, fo is LG to GH, that is, the Sides about the equal Angles EBC, LGH are proportional. Wherefore the Triangle EBC is equiangular to the Triangle LGH; and confequently alfo fimilar to it. For the fame Reason, the Triangle ECD, is likewife fimilar to the Triangle LHK; therefore the fimilar Polygons ABCDE, FGHKL, are divided into equal Numbers of fimilar Triangles.

I fay, they are alfo homologous to the Wholes, that is, that the Triangles are proportional; and the Antecedents are ABE, EBC, ECD, and their Confequents FGL, LGH, LHK. And the Polygon ABCDE, to the Polygon FGHKL, is in the duplicate Proportion of an homologous Side of the one, to an homologous Side of the other, that is, AB to FC.

*

For because the Triangle ABE is fimilar to the Triangle F GL, the Triangle ABE, shall be* to the 19 of this. Triangle F GL, in the duplicate Proportion of BE to GL: For the fame Reason, the Triangle BEC, to the Triangle GLH, is* in a duplicate Proportion of BE to GL: Therefore the Triangle ABE ist tot 11. 5. the Triangle F GL, as the Triangle BEC is to the Triangle GLH. Again, because the Triangle EBC is fimilar to the Triangle LGH; the Triangle EBC to the Triangle LGH, fhall be in the duplicate Proportion of the Right Line CE to the Right Line HL; and so likewife the Triangle ECD to the Tri

A

12. 5.

F

BGX

angle LHK, fhall be in the duplicate Proportion of CE to HL. Therefore the Triangle BEC is to the Triangle LGH, as the Triangle CED is to the Triangle LHK. But it has been proved, that the Triangle EBC is to the Triangle LGH, as the Triangle ABE is to the Triangle FGL: Therefore as the Triangle ABE is to the Triangle F GL, fo is the Triangle BEC to the Triangle GHL; and fo is the Triangle ECD to the Triangle LHK. But as one of the Antecedents is to one of the Confequents, fo are all the Antecedents to all the Confequents. Wherefore as the Triangle ABE is to the Triangle F GL, fo is the Polygon ABCDE to the Polygon FGHKL: But the Triangle ABE to the Triangle F GL, 'is in the duplicate Proportion of the homologous Side AB to the homologous Side F G; for fimilar Triangles are in the duplicate Proportion of the homologous Sides. Wherefore the Polygon ABCDE, to the Polygon FGHKL, is in the duplicate Proportion of the homologous Side AB to the homologous Side FG. Therefore fimilar Polygons are divided into fimilar Triangles, equal in Number, and homologous to the Wholes; and Polygon to Polygon, is in the duplicate Proportion of that which one homologous Side has to the other; which was to be demonftrated.

It may be demonftrated after the fame manner that fimilar quadrilateral Figures are to each other in the duplicate Proportion of their homologous Sides; and this has been already proved in Triangles.

Coroll. 1. Therefore univerfally fimilar Right-lin❜d Figures, are to one another in the duplicate Proportion of their homologous Sides; and if X be taken a third Proportional to AB and FG, then AB will have to X a duplicate Proportion of that which AB has to F G ; and a Polygon to a Polygon, and a quadrilateral 'Figure to a quadrilateral Figure, will be in the duplicate Proportion of that which one homologous fide has to the other; that is, AB to FG; but this has been proved in Triangles.

2. Therefore univerfally it is manifeft, if three Right Lines be proportional, as the firft is to the third, fo is a Figure defcribed upon the firft, to a fimilar and

fimilarly